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What is an example of a separable, simple $C^{\ast}$-algebra that admits two different tracial states?

EDIT: Julien has pointed to a number of avenues to answer this question. If anyone has an electronic copy of the paper of Longo he links to in the comment below, please post a summary of the argument. It would be nice to have access to a nice simple construction as advertised in the abstract of that paper.

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  • $\begingroup$ Have you considered using classification theorems, e.g. those for AF- or AI-algebras? $\endgroup$ – Michael Jul 5 '13 at 22:26
  • $\begingroup$ Or, if you really want to use a big-hammer, try the classification theorem for tracially AF algebras by Lin. You can find the outlines of these results in Rordam's half-volume on classification of nuclear C*-algebras. $\endgroup$ – Michael Jul 5 '13 at 22:33
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    $\begingroup$ Effros-Handelman-Shen proved that every countable Riesz group arises as the dimension- $K_0$ group of an AF-algebra. And the corresponding AF-algebra is simple if and only if the $K_0$ group has no nontrivial order ideal. So it amounts to taking such acountable Riesz group, with at least two distinct states (positive homomorphisms in $(\mathbb{R},\mathbb{R}^+,[0,1])$). $\endgroup$ – Julien Jul 6 '13 at 1:03
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    $\begingroup$ To be fair, one should probably add that Blackadar and Goodearl (independently, and prior to EHS) proved that every metrizable Choquet simplex arises as the tracial state space (up to an affine homeomorphism) of a simple unital AF-algebra. Note also that the tracial state space is a part of Elliott's invariant which truly plays a role in classification for some more complicated classes than AF-algebras. $\endgroup$ – Julien Jul 6 '13 at 18:09
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    $\begingroup$ On the crossed product side, $C(X)\rtimes_\phi \mathbb{Z}$ is simple if and only if the homeomorphism $\phi$ of the infinite compact Hausdorff space $X$ is minimal. Traces correspond to $\phi$-invariant probability measures on $X$. I don't have access to this paper of Longo, but it seems to address your question (provided the algebras are indeed separable). $\endgroup$ – Julien Jul 6 '13 at 18:41
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Longo's example is as follows: he considers a C$^*$-dynamical system $(A,G,\alpha)$ with $A$ unital, simple, and with unique trace $\tau$; and $G$ discrete abelian.

He notes that in this situation $\tau$ is $\alpha$-invariant (by the uniqueness of the trace) and thus $\alpha$ extends to $\bar\alpha:G\to\mbox{Aut}(\pi_\tau(A)'')$ (I didn't think why this is true).

And then he considers an additional condition on the action of the group. He requires the existence of nonzero $t\in G$ such that $\alpha_t$ is not inner, but $\bar\alpha_t=v\cdot v^*$, where $v\in\pi_\tau(A)''$ is an $\bar\alpha$-invariant unitary.

Under those conditions, he proves that $A\rtimes_\alpha G$ is simple and has at least two traces.

As a concrete example, he considers $A=\bigotimes_{n=1}^\infty\,M_{2^n}(\mathbb C)$, $G=\mathbb Z_2$, and the nontrivial element of $\mathbb Z_2$ given by conjugation by $\bigotimes_nu_n$, where $u_n$ is the diagonal matrix on $M_{2^n}(\mathbb C)$ that has diagonal $1,\ldots,1,-1$.

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    $\begingroup$ Thanks and +1 in the name of non-university people like me who don't have access to the locked knowledge. $\endgroup$ – Julien Jul 18 '13 at 3:58
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Blackadar showed [Traces on simple AF C*-algebras. J. Funct. Anal. 38 (1980), no. 2, 156–168] that every metrizable Choquet simplex arises as the trace space of a simple, AF algebra.

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  • $\begingroup$ Is it correct to say that Goodearl, as I mentioned above, proved that as well independently? I've read that somewhere (a survey by Rordam I think). But I don't know the reference. $\endgroup$ – Julien Jul 18 '13 at 3:55
  • $\begingroup$ julien, you're right. Actually the result is (as Blackadar states in the referenced paper) just a corollary to a result of Goodearl (I should have read Blackadar's proof before answering) $\endgroup$ – user85775 Jul 19 '13 at 4:10

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